Possible values of A,B,C?

2A+3B+5C=19


A,B,C could be at least 1








what are the possible values of A,B,C which would satisfy the above?














whats the scientific way to find it out ?





or do i have to start with random guess ? starting with random guess is painful. i want to find out the answer in quick steps.





any scientific steps to follow ?

Possible values of A,B,C?
Well, you know C%26lt;4 to start with.





Also, since the total sum is odd, either B or C must be odd, but not both. Since C has the fewest possible values, I'd start with that.





If C=3, 5C = 15 and there is no combination of 2A + 3B (where A and B must be 1 or more) that could equal 4. So C must be 1 or 2.





If C=2, 2A + 3B must equal 9. So B must be 3 or less. If B = 3, A = 0, which is not acceptable. B must be odd (see above paragraph), so B must be 1 (yielding A = 3).





If C = 1, 2A + 3B must equal 14. So B must be 1, 2, 3 or 4. As in the paragraph above, if C is odd, B must be even. So if B = 4, 2A + 12 + 5 = 19, so A = 1. If B = 2, 2A + 6 + 5 = 19, so A = 4.





We now have 3 solutions:


I) A = 3, B = 1, C = 2


II) A = 1, B = 4, C = 1


III) A = 4, B = 2, C = 1
Reply:u can use a graphing calculator
Reply:A = 3


B = 1


C = 2
Reply:I think you are suppose to come up with you own values to satisfy the equation





2A+3B+5C=19


2(0)+3(3)+5(2)=19


0+9+10=19


19+19





therefore A=-2, B=3, C=2
Reply:If you have three variables, then with only one equation constraining the values there will be infinitely many answers. For exactly one answer for each, you would need three equations.





Perhaps the easiest way is to create a systematic list of the three values and watch for patterns to develop.
Reply:A=3


B=1


C=2


2*3+3*1+5*2=19


Other answers do exist and are equally valid.


The scientific method, using a graphing calculator, the unscientific method, take the highest number and assign a number to the variable (C) so the remainder can be made with additions of the lower two numbers. 9 = 3 + 2 + 2 + 2
Reply:if the minimum value of each variable is 1, then set two variables to 1 and solve for the third. like this If b is one and c is one then a would be 11/2. then do that for the other 2 variables. then you will know the minimum and maximum value of each variable.
Reply:For A,B, and C where they can equal any rational number including factions/decimals,





Where


A=(19-3B-5C)/2


B=(19-2A-5C)/3


C=(19-2A-3B)/5





for any value of two of the three variables, the third will equal some absolute value.





If on the other hand the values of A,B, and C must be whole numbers, then a possible combination is A=1, B=-1, C=4, thus giving you 2(1) + 3(-1) +5(4) =19





Or if the values of A,B, and C must be non-negative


A=8, B=1, and C=0; 2(8) + 3(1) + 5(0) = 19





Or if the values must be non-negative and non-zero


A=4, B=2, C=1





Your question is good, however, I solved this equation 4 ways in under 5 minutes. Ultimately it was guess and check, but it wasnt assigning values immediately to A,B,and C, it was more of a task of subtracting the value of 2,3, and 5 from 19 until you reach 0. For instance, 19-5-3-2-2-2-2-3=0
Reply:1 euqation with 3 unknowns...doesnt work





Thats the end of that.

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